How far did the car travel in the fourth second?

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The car accelerates from 14 m/s to 26 m/s over 4 seconds, resulting in an acceleration of 3 m/s². The distance traveled in the first four seconds is calculated to be 80 meters. However, the question specifically asks for the distance traveled during the fourth second, which is not the same as the total distance. The correct method involves calculating the distance covered in just that second, leading to the answer of 24.5 meters. Clarification on the distinction between total distance and distance in a specific second is crucial for solving the problem correctly.
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Homework Statement


A car accelerates from 14 m/s to 26m/s in 4s. How far did the car travel in the fourth second?


Homework Equations


a=vf-vi/t = 26-14/4=3

d=vf^2-vi^2/2a=26^2-14^2/3=80


The answer is 24.5, but I keep getting 80 and I don't know why.
 
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Your answer is correct, perhaps the book is wrong?
 


You've worked out the distance traveled in four seconds, the question is asking for the distance traveled in the fourth second.

If you're still stuck;

Try using the equation d = vi * t + (0.5 * a * t2)
Use it to work out how far the car travels in three seconds, then how far it travels in four seconds (your value of 80m is correct).
Then use the two to work out how far it traveled in the fourth second.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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